Q.A light rigid rod $3\ \text{m}$ long is pivoted at its midpoint and can turn freely (like a see-saw). A weight of $20\ \text{N}$ hangs at a distance of $1.2\ \text{m}$ from the pivot on one side. Find the weight that must hang at a distance of $0.8\ \text{m}$ from the pivot on the other side to keep the rod exactly balanced.

Consider a light (effectively massless) rigid rod pivoted at some point along its length, with two forces $F_1$ and $F_2$ acting perpendicular to the rod at distances $d_1$ and $d_2$ respectively from the pivot, plus the pivot's own normal reaction force $N$. For the rod to remain in equilibrium, both the net force and the net torque about the pivot must vanish.

Net force zero: $N=F_1+F_2$ (the pivot's upward reaction balances the two downward forces).

Net torque zero about the pivot (the reaction force $N$ itself passes through the pivot and so contributes no torque there):

This is the principle of moments: for a rod (or lever, or beam balance) in rotational equilibrium about its pivot, the moment (force × distance) on one side must exactly equal the moment on the other side. Rearranging, $\dfrac{F_1}{F_2}=\dfrac{d_2}{d_1}$. …